JEE Exam > JEE Questions > The number of possible natural oscillations o...
Start Learning for Free
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)
Correct answer is '6'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The number of possible natural oscillations of air column in a pipe cl...
Frequency COP, nN = (2N + 1) v/4L
for N = 0, n0 = 100 Hz
n = 1, n1 = 300 Hz
n = 2, n2 = 500 Hz
n = 3, n3 = 700 Hz
n = 4, n4 = 900 Hz
n = 5, n5 = 1100 Hz
Total = 6
Which are less than 1250 Hz.
for N = 0, n0 = 100 Hz
n = 1, n1 = 300 Hz
n = 2, n2 = 500 Hz
n = 3, n3 = 700 Hz
n = 4, n4 = 900 Hz
n = 5, n5 = 1100 Hz
Total = 6
Which are less than 1250 Hz.
Most Upvoted Answer
The number of possible natural oscillations of air column in a pipe cl...
To find the number of possible natural oscillations of an air column in a pipe closed at one end, we need to consider the fundamental frequency and the overtones.
The fundamental frequency is the lowest frequency at which the air column can vibrate, and it is given by the formula:
f1 = v / 4L
where f1 is the fundamental frequency, v is the velocity of sound, and L is the length of the air column.
In this case, the length of the air column is 85 cm, which is equal to 0.85 m. The velocity of sound is given as 340 m/s. Plugging these values into the formula, we can find the fundamental frequency:
f1 = 340 / (4 * 0.85) = 100 Hz
The fundamental frequency is 100 Hz.
The overtones are frequencies that are multiples of the fundamental frequency. The nth overtone can be found using the formula:
fn = nf1
where fn is the frequency of the nth overtone, n is the overtone number, and f1 is the fundamental frequency.
To find the number of possible natural oscillations below 1250 Hz, we need to find the highest overtone frequency that is below 1250 Hz. We can do this by dividing 1250 by the fundamental frequency:
1250 / 100 = 12.5
So the highest overtone frequency below 1250 Hz is the 12th overtone. Therefore, there are 12 possible natural oscillations of the air column in the pipe closed at one end whose frequencies lie below 1250 Hz.
The fundamental frequency is the lowest frequency at which the air column can vibrate, and it is given by the formula:
f1 = v / 4L
where f1 is the fundamental frequency, v is the velocity of sound, and L is the length of the air column.
In this case, the length of the air column is 85 cm, which is equal to 0.85 m. The velocity of sound is given as 340 m/s. Plugging these values into the formula, we can find the fundamental frequency:
f1 = 340 / (4 * 0.85) = 100 Hz
The fundamental frequency is 100 Hz.
The overtones are frequencies that are multiples of the fundamental frequency. The nth overtone can be found using the formula:
fn = nf1
where fn is the frequency of the nth overtone, n is the overtone number, and f1 is the fundamental frequency.
To find the number of possible natural oscillations below 1250 Hz, we need to find the highest overtone frequency that is below 1250 Hz. We can do this by dividing 1250 by the fundamental frequency:
1250 / 100 = 12.5
So the highest overtone frequency below 1250 Hz is the 12th overtone. Therefore, there are 12 possible natural oscillations of the air column in the pipe closed at one end whose frequencies lie below 1250 Hz.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer?
Question Description
The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer?.
The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer?.
Solutions for The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer?, a detailed solution for The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? has been provided alongside types of The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The number of possible natural oscillations of air column in a pipe closed at one end of length85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms–1)Correct answer is '6'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup